Prove

Question:

$\sin ^{3} x \cos ^{3} x$

Solution:

Let $\begin{aligned} I &=\int \sin ^{3} x \cos ^{3} x \cdot d x \\ &=\int \cos ^{3} x \cdot \sin ^{2} x \cdot \sin x \cdot d x \\ &=\int \cos ^{3} x\left(1-\cos ^{2} x\right) \sin x \cdot d x \end{aligned}$

Let $\cos x=t$

$\Rightarrow-\sin x \cdot d x=d t$

$\Rightarrow I=-\int t^{3}\left(1-t^{2}\right) d t$

$=-\int\left(t^{3}-t^{5}\right) d t$

$=-\left\{\frac{t^{4}}{4}-\frac{t^{6}}{6}\right\}+\mathrm{C}$

$=-\left\{\frac{\cos ^{4} x}{4}-\frac{\cos ^{6} x}{6}\right\}+\mathrm{C}$

$=\frac{\cos ^{6} x}{6}-\frac{\cos ^{4} x}{4}+\mathrm{C}$

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